On the integral $int_{0}^{1/2}frac{text{Li}_3(1-z)}{sqrt{z(1-z)}},dz$
up vote
8
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This questions is related to my previous one.
I am interested in a explicit evaluation in terms of Euler sums for
$$ int_{0}^{pi/4}text{Li}_3(cos^2theta),dtheta = frac{1}{2}int_{0}^{1/2}frac{text{Li}_3(1-z)}{sqrt{z(1-z)}},dz.$$
It is not difficult to show that
$$ int_{0}^{color{red}{1}}frac{text{Li}_3(1-z)}{sqrt{z(1-z)}},dz =-frac{pi^3}{3}log(2)+frac{4pi}{3}log^3(2)+2pizeta(3)tag{A}$$
but I have not managed to make a wise use of the trilogarithm functional identities for computing
$$ int_{0}^{1/2}frac{text{Li}_3(z)}{sqrt{z(1-z)}},dzstackrel{text{IBP}}{longrightarrow}int_{0}^{pi/4}thetacot(theta)text{Li}_2(sin^2theta),dtheta quadtext{or}quadint_{0}^{1/2}frac{text{Li}_3(z)-text{Li}_3(1-z)}{sqrt{z(1-z)}},dz ,$$
which would have solved the problem. One might need substantial extensions of the result about $mathcal{I}(a,b)$ proved here by nospoon. I am expecting the integral above to be related with Euler sums with (total) weight five. Maybe the shifted-Fourier-Chebyshev expansion of $text{Li}_3(x)$ over $(0,1)$ is already known in the literature, but I have not been able to find it.
integration reference-request special-functions hypergeometric-function polylogarithm
asked May 28 at 23:55
Jack D'Aurizio
283k33275653
add a comment |
up vote
8
down vote
favorite
This questions is related to my previous one.
I am interested in a explicit evaluation in terms of Euler sums for
$$ int_{0}^{pi/4}text{Li}_3(cos^2theta),dtheta = frac{1}{2}int_{0}^{1/2}frac{text{Li}_3(1-z)}{sqrt{z(1-z)}},dz.$$
It is not difficult to show that
$$ int_{0}^{color{red}{1}}frac{text{Li}_3(1-z)}{sqrt{z(1-z)}},dz =-frac{pi^3}{3}log(2)+frac{4pi}{3}log^3(2)+2pizeta(3)tag{A}$$
but I have not managed to make a wise use of the trilogarithm functional identities for computing
$$ int_{0}^{1/2}frac{text{Li}_3(z)}{sqrt{z(1-z)}},dzstackrel{text{IBP}}{longrightarrow}int_{0}^{pi/4}thetacot(theta)text{Li}_2(sin^2theta),dtheta quadtext{or}quadint_{0}^{1/2}frac{text{Li}_3(z)-text{Li}_3(1-z)}{sqrt{z(1-z)}},dz ,$$
which would have solved the problem. One might need substantial extensions of the result about $mathcal{I}(a,b)$ proved here by nospoon. I am expecting the integral above to be related with Euler sums with (total) weight five. Maybe the shifted-Fourier-Chebyshev expansion of $text{Li}_3(x)$ over $(0,1)$ is already known in the literature, but I have not been able to find it.
integration reference-request special-functions hypergeometric-function polylogarithm
asked May 28 at 23:55
Jack D'Aurizio
283k33275653
There is no tag "polylogarithms"? (I cannot understand the classification in (hypergeometric-function)...)
– dan_fulea
May 29 at 0:15
@dan_fulea: You are right, I forgot to add it. The hypergeometric tag is due to the fact that such integral gives a closed form for a peculiar $phantom{}_5 F_4$ with half-integer parameters.
– Jack D'Aurizio
May 29 at 0:16
add a comment |
up vote
8
down vote
favorite
up vote
8
down vote
favorite
This questions is related to my previous one.
I am interested in a explicit evaluation in terms of Euler sums for
$$ int_{0}^{pi/4}text{Li}_3(cos^2theta),dtheta = frac{1}{2}int_{0}^{1/2}frac{text{Li}_3(1-z)}{sqrt{z(1-z)}},dz.$$
It is not difficult to show that
$$ int_{0}^{color{red}{1}}frac{text{Li}_3(1-z)}{sqrt{z(1-z)}},dz =-frac{pi^3}{3}log(2)+frac{4pi}{3}log^3(2)+2pizeta(3)tag{A}$$
but I have not managed to make a wise use of the trilogarithm functional identities for computing
$$ int_{0}^{1/2}frac{text{Li}_3(z)}{sqrt{z(1-z)}},dzstackrel{text{IBP}}{longrightarrow}int_{0}^{pi/4}thetacot(theta)text{Li}_2(sin^2theta),dtheta quadtext{or}quadint_{0}^{1/2}frac{text{Li}_3(z)-text{Li}_3(1-z)}{sqrt{z(1-z)}},dz ,$$
which would have solved the problem. One might need substantial extensions of the result about $mathcal{I}(a,b)$ proved here by nospoon. I am expecting the integral above to be related with Euler sums with (total) weight five. Maybe the shifted-Fourier-Chebyshev expansion of $text{Li}_3(x)$ over $(0,1)$ is already known in the literature, but I have not been able to find it.
integration reference-request special-functions hypergeometric-function polylogarithm
asked May 28 at 23:55
Jack D'Aurizio
283k33275653
This questions is related to my previous one.
I am interested in a explicit evaluation in terms of Euler sums for
$$ int_{0}^{pi/4}text{Li}_3(cos^2theta),dtheta = frac{1}{2}int_{0}^{1/2}frac{text{Li}_3(1-z)}{sqrt{z(1-z)}},dz.$$
It is not difficult to show that
$$ int_{0}^{color{red}{1}}frac{text{Li}_3(1-z)}{sqrt{z(1-z)}},dz =-frac{pi^3}{3}log(2)+frac{4pi}{3}log^3(2)+2pizeta(3)tag{A}$$
but I have not managed to make a wise use of the trilogarithm functional identities for computing
$$ int_{0}^{1/2}frac{text{Li}_3(z)}{sqrt{z(1-z)}},dzstackrel{text{IBP}}{longrightarrow}int_{0}^{pi/4}thetacot(theta)text{Li}_2(sin^2theta),dtheta quadtext{or}quadint_{0}^{1/2}frac{text{Li}_3(z)-text{Li}_3(1-z)}{sqrt{z(1-z)}},dz ,$$
which would have solved the problem. One might need substantial extensions of the result about $mathcal{I}(a,b)$ proved here by nospoon. I am expecting the integral above to be related with Euler sums with (total) weight five. Maybe the shifted-Fourier-Chebyshev expansion of $text{Li}_3(x)$ over $(0,1)$ is already known in the literature, but I have not been able to find it.
integration reference-request special-functions hypergeometric-function polylogarithm
integration reference-request special-functions hypergeometric-function polylogarithm
asked May 28 at 23:55
Jack D'Aurizio
283k33275653
asked May 28 at 23:55
Jack D'Aurizio
283k33275653
edited May 29 at 0:15
asked May 28 at 23:55
Jack D'Aurizio
283k33275653
asked May 28 at 23:55
Jack D'Aurizio
283k33275653
asked May 28 at 23:55
Jack D'Aurizio
283k33275653
283k33275653
There is no tag "polylogarithms"? (I cannot understand the classification in (hypergeometric-function)...)
– dan_fulea
May 29 at 0:15
@dan_fulea: You are right, I forgot to add it. The hypergeometric tag is due to the fact that such integral gives a closed form for a peculiar $phantom{}_5 F_4$ with half-integer parameters.
– Jack D'Aurizio
May 29 at 0:16
add a comment |
There is no tag "polylogarithms"? (I cannot understand the classification in (hypergeometric-function)...)
– dan_fulea
May 29 at 0:15
@dan_fulea: You are right, I forgot to add it. The hypergeometric tag is due to the fact that such integral gives a closed form for a peculiar $phantom{}_5 F_4$ with half-integer parameters.
– Jack D'Aurizio
May 29 at 0:16
There is no tag "polylogarithms"? (I cannot understand the classification in (hypergeometric-function)...)
– dan_fulea
May 29 at 0:15
There is no tag "polylogarithms"? (I cannot understand the classification in (hypergeometric-function)...)
– dan_fulea
May 29 at 0:15
@dan_fulea: You are right, I forgot to add it. The hypergeometric tag is due to the fact that such integral gives a closed form for a peculiar $phantom{}_5 F_4$ with half-integer parameters.
– Jack D'Aurizio
May 29 at 0:16
@dan_fulea: You are right, I forgot to add it. The hypergeometric tag is due to the fact that such integral gives a closed form for a peculiar $phantom{}_5 F_4$ with half-integer parameters.
– Jack D'Aurizio
May 29 at 0:16
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
This is meant to be just a comment (too large to add it in the comments section)
$$I=operatorname{Li}_4left(frac{1}{2}right)+frac{2}{3}sum_{n=2}^{infty}(-1)^{n-1}frac{n(overbrace{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)})}^{G1}-3 (overbrace{H_n^2+ H_n^{(2)})}^{G2}}{2n(2n-1)(2n-2)}.$$
A wise next move would be to split the series into 2 series using the groups $G1$ and $G2$ and try to calculate the resulting series like that. In the mathematical literature there are nice, useful representations for those harmonic numbers groups.
answered May 29 at 15:33
user 1357113
22.2k875224
1
The series related to G2 are manageable. Then we may perform a partial fraction decomposition and exploit the results of Xu, Yang and Zhang (Explicit evaluation of quadratic Euler sums) to compute two other pieces. The hardcore part is the evaluation of $$ sum_{ngeq 0}(-1)^n frac{H_n^3+3H_n H_n^{(2)}+2H_n^{(3)}}{2n-1} $$ which does not seem to be covered by classical results.
– Jack D'Aurizio
May 29 at 16:29
On the other hand such sum is related to $int_{0}^{pi/4}log^3(costheta),dtheta$ which has been computed by nospoon. Can you provide the derivation of the identity above in full detail? I believe I might be able to turn it into a complete solution. (+1) in the meanwhile.
– Jack D'Aurizio
May 29 at 16:31
1
@JackD'Aurizio I checked all the integrals behind the scene. I think all of them can be calculated by combining real and complex methods, but there is some work to do there. Before making any update I want to investigate the possibility of getting major shortcuts.
– user 1357113
May 29 at 17:25
Perfectly fine. You are probably already aware of it, but consider that the integrals $int_{0}^{pi/4}log^mu(cos t),dt$ and $int_{0}^{pi/4}log^mu(sin t),dt$ for $muin{0,1,2,3}$ are known, even if WA does not return them in a nice closed form.
– Jack D'Aurizio
May 29 at 17:57
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
This is meant to be just a comment (too large to add it in the comments section)
$$I=operatorname{Li}_4left(frac{1}{2}right)+frac{2}{3}sum_{n=2}^{infty}(-1)^{n-1}frac{n(overbrace{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)})}^{G1}-3 (overbrace{H_n^2+ H_n^{(2)})}^{G2}}{2n(2n-1)(2n-2)}.$$
A wise next move would be to split the series into 2 series using the groups $G1$ and $G2$ and try to calculate the resulting series like that. In the mathematical literature there are nice, useful representations for those harmonic numbers groups.
answered May 29 at 15:33
user 1357113
22.2k875224
1
The series related to G2 are manageable. Then we may perform a partial fraction decomposition and exploit the results of Xu, Yang and Zhang (Explicit evaluation of quadratic Euler sums) to compute two other pieces. The hardcore part is the evaluation of $$ sum_{ngeq 0}(-1)^n frac{H_n^3+3H_n H_n^{(2)}+2H_n^{(3)}}{2n-1} $$ which does not seem to be covered by classical results.
– Jack D'Aurizio
May 29 at 16:29
On the other hand such sum is related to $int_{0}^{pi/4}log^3(costheta),dtheta$ which has been computed by nospoon. Can you provide the derivation of the identity above in full detail? I believe I might be able to turn it into a complete solution. (+1) in the meanwhile.
– Jack D'Aurizio
May 29 at 16:31
1
@JackD'Aurizio I checked all the integrals behind the scene. I think all of them can be calculated by combining real and complex methods, but there is some work to do there. Before making any update I want to investigate the possibility of getting major shortcuts.
– user 1357113
May 29 at 17:25
Perfectly fine. You are probably already aware of it, but consider that the integrals $int_{0}^{pi/4}log^mu(cos t),dt$ and $int_{0}^{pi/4}log^mu(sin t),dt$ for $muin{0,1,2,3}$ are known, even if WA does not return them in a nice closed form.
– Jack D'Aurizio
May 29 at 17:57
add a comment |
up vote
3
down vote
This is meant to be just a comment (too large to add it in the comments section)
$$I=operatorname{Li}_4left(frac{1}{2}right)+frac{2}{3}sum_{n=2}^{infty}(-1)^{n-1}frac{n(overbrace{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)})}^{G1}-3 (overbrace{H_n^2+ H_n^{(2)})}^{G2}}{2n(2n-1)(2n-2)}.$$
A wise next move would be to split the series into 2 series using the groups $G1$ and $G2$ and try to calculate the resulting series like that. In the mathematical literature there are nice, useful representations for those harmonic numbers groups.
answered May 29 at 15:33
user 1357113
22.2k875224
1
The series related to G2 are manageable. Then we may perform a partial fraction decomposition and exploit the results of Xu, Yang and Zhang (Explicit evaluation of quadratic Euler sums) to compute two other pieces. The hardcore part is the evaluation of $$ sum_{ngeq 0}(-1)^n frac{H_n^3+3H_n H_n^{(2)}+2H_n^{(3)}}{2n-1} $$ which does not seem to be covered by classical results.
– Jack D'Aurizio
May 29 at 16:29
On the other hand such sum is related to $int_{0}^{pi/4}log^3(costheta),dtheta$ which has been computed by nospoon. Can you provide the derivation of the identity above in full detail? I believe I might be able to turn it into a complete solution. (+1) in the meanwhile.
– Jack D'Aurizio
May 29 at 16:31
1
@JackD'Aurizio I checked all the integrals behind the scene. I think all of them can be calculated by combining real and complex methods, but there is some work to do there. Before making any update I want to investigate the possibility of getting major shortcuts.
– user 1357113
May 29 at 17:25
Perfectly fine. You are probably already aware of it, but consider that the integrals $int_{0}^{pi/4}log^mu(cos t),dt$ and $int_{0}^{pi/4}log^mu(sin t),dt$ for $muin{0,1,2,3}$ are known, even if WA does not return them in a nice closed form.
– Jack D'Aurizio
May 29 at 17:57
add a comment |
up vote
3
down vote
up vote
3
down vote
This is meant to be just a comment (too large to add it in the comments section)
$$I=operatorname{Li}_4left(frac{1}{2}right)+frac{2}{3}sum_{n=2}^{infty}(-1)^{n-1}frac{n(overbrace{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)})}^{G1}-3 (overbrace{H_n^2+ H_n^{(2)})}^{G2}}{2n(2n-1)(2n-2)}.$$
A wise next move would be to split the series into 2 series using the groups $G1$ and $G2$ and try to calculate the resulting series like that. In the mathematical literature there are nice, useful representations for those harmonic numbers groups.
answered May 29 at 15:33
user 1357113
22.2k875224
This is meant to be just a comment (too large to add it in the comments section)
$$I=operatorname{Li}_4left(frac{1}{2}right)+frac{2}{3}sum_{n=2}^{infty}(-1)^{n-1}frac{n(overbrace{H_n^3+3H_nH_n^{(2)}+2H_n^{(3)})}^{G1}-3 (overbrace{H_n^2+ H_n^{(2)})}^{G2}}{2n(2n-1)(2n-2)}.$$
A wise next move would be to split the series into 2 series using the groups $G1$ and $G2$ and try to calculate the resulting series like that. In the mathematical literature there are nice, useful representations for those harmonic numbers groups.
answered May 29 at 15:33
user 1357113
22.2k875224
answered May 29 at 15:33
user 1357113
22.2k875224
answered May 29 at 15:33
user 1357113
22.2k875224
answered May 29 at 15:33
user 1357113
22.2k875224
22.2k875224
1
The series related to G2 are manageable. Then we may perform a partial fraction decomposition and exploit the results of Xu, Yang and Zhang (Explicit evaluation of quadratic Euler sums) to compute two other pieces. The hardcore part is the evaluation of $$ sum_{ngeq 0}(-1)^n frac{H_n^3+3H_n H_n^{(2)}+2H_n^{(3)}}{2n-1} $$ which does not seem to be covered by classical results.
– Jack D'Aurizio
May 29 at 16:29
On the other hand such sum is related to $int_{0}^{pi/4}log^3(costheta),dtheta$ which has been computed by nospoon. Can you provide the derivation of the identity above in full detail? I believe I might be able to turn it into a complete solution. (+1) in the meanwhile.
– Jack D'Aurizio
May 29 at 16:31
1
@JackD'Aurizio I checked all the integrals behind the scene. I think all of them can be calculated by combining real and complex methods, but there is some work to do there. Before making any update I want to investigate the possibility of getting major shortcuts.
– user 1357113
May 29 at 17:25
Perfectly fine. You are probably already aware of it, but consider that the integrals $int_{0}^{pi/4}log^mu(cos t),dt$ and $int_{0}^{pi/4}log^mu(sin t),dt$ for $muin{0,1,2,3}$ are known, even if WA does not return them in a nice closed form.
– Jack D'Aurizio
May 29 at 17:57
add a comment |
1
The series related to G2 are manageable. Then we may perform a partial fraction decomposition and exploit the results of Xu, Yang and Zhang (Explicit evaluation of quadratic Euler sums) to compute two other pieces. The hardcore part is the evaluation of $$ sum_{ngeq 0}(-1)^n frac{H_n^3+3H_n H_n^{(2)}+2H_n^{(3)}}{2n-1} $$ which does not seem to be covered by classical results.
– Jack D'Aurizio
May 29 at 16:29
On the other hand such sum is related to $int_{0}^{pi/4}log^3(costheta),dtheta$ which has been computed by nospoon. Can you provide the derivation of the identity above in full detail? I believe I might be able to turn it into a complete solution. (+1) in the meanwhile.
– Jack D'Aurizio
May 29 at 16:31
1
@JackD'Aurizio I checked all the integrals behind the scene. I think all of them can be calculated by combining real and complex methods, but there is some work to do there. Before making any update I want to investigate the possibility of getting major shortcuts.
– user 1357113
May 29 at 17:25
Perfectly fine. You are probably already aware of it, but consider that the integrals $int_{0}^{pi/4}log^mu(cos t),dt$ and $int_{0}^{pi/4}log^mu(sin t),dt$ for $muin{0,1,2,3}$ are known, even if WA does not return them in a nice closed form.
– Jack D'Aurizio
May 29 at 17:57
1
1
The series related to G2 are manageable. Then we may perform a partial fraction decomposition and exploit the results of Xu, Yang and Zhang (Explicit evaluation of quadratic Euler sums) to compute two other pieces. The hardcore part is the evaluation of $$ sum_{ngeq 0}(-1)^n frac{H_n^3+3H_n H_n^{(2)}+2H_n^{(3)}}{2n-1} $$ which does not seem to be covered by classical results.
– Jack D'Aurizio
May 29 at 16:29
The series related to G2 are manageable. Then we may perform a partial fraction decomposition and exploit the results of Xu, Yang and Zhang (Explicit evaluation of quadratic Euler sums) to compute two other pieces. The hardcore part is the evaluation of $$ sum_{ngeq 0}(-1)^n frac{H_n^3+3H_n H_n^{(2)}+2H_n^{(3)}}{2n-1} $$ which does not seem to be covered by classical results.
– Jack D'Aurizio
May 29 at 16:29
On the other hand such sum is related to $int_{0}^{pi/4}log^3(costheta),dtheta$ which has been computed by nospoon. Can you provide the derivation of the identity above in full detail? I believe I might be able to turn it into a complete solution. (+1) in the meanwhile.
– Jack D'Aurizio
May 29 at 16:31
On the other hand such sum is related to $int_{0}^{pi/4}log^3(costheta),dtheta$ which has been computed by nospoon. Can you provide the derivation of the identity above in full detail? I believe I might be able to turn it into a complete solution. (+1) in the meanwhile.
– Jack D'Aurizio
May 29 at 16:31
1
1
@JackD'Aurizio I checked all the integrals behind the scene. I think all of them can be calculated by combining real and complex methods, but there is some work to do there. Before making any update I want to investigate the possibility of getting major shortcuts.
– user 1357113
May 29 at 17:25
@JackD'Aurizio I checked all the integrals behind the scene. I think all of them can be calculated by combining real and complex methods, but there is some work to do there. Before making any update I want to investigate the possibility of getting major shortcuts.
– user 1357113
May 29 at 17:25
Perfectly fine. You are probably already aware of it, but consider that the integrals $int_{0}^{pi/4}log^mu(cos t),dt$ and $int_{0}^{pi/4}log^mu(sin t),dt$ for $muin{0,1,2,3}$ are known, even if WA does not return them in a nice closed form.
– Jack D'Aurizio
May 29 at 17:57
Perfectly fine. You are probably already aware of it, but consider that the integrals $int_{0}^{pi/4}log^mu(cos t),dt$ and $int_{0}^{pi/4}log^mu(sin t),dt$ for $muin{0,1,2,3}$ are known, even if WA does not return them in a nice closed form.
– Jack D'Aurizio
May 29 at 17:57
add a comment |
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There is no tag "polylogarithms"? (I cannot understand the classification in (hypergeometric-function)...)
– dan_fulea
May 29 at 0:15
@dan_fulea: You are right, I forgot to add it. The hypergeometric tag is due to the fact that such integral gives a closed form for a peculiar $phantom{}_5 F_4$ with half-integer parameters.
– Jack D'Aurizio
May 29 at 0:16